Motion – many examples surround us an ice skater coasting

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Transcript Motion – many examples surround us an ice skater coasting

Motion – many examples surround us
an ice skater coasting along
a car screeching to a halt
a ball dropped from the hand
a feather floating in the wind
a shell fired from a canon
Of all these, simplest to analyze:
ice skater coasting
Useful Physical Quantities to Describe Motion
Position x(t), as a function of time
- describes the position of an object at any time.
- a vector quantity (w/ magnitude, direction)
- in metric units of meters
Velocity v(t)
- describes how position x(t) changes in time.
- a vector quantity measured in units of meters/second
v(t) = x/t
- magnitude of velocity is called ‘speed’
Acceleration a(t)
- describes how velocity v(t) changes in time.
- a vector quantity measured in units of m/s2
a(t) = v/ t
Different Types of Motion
Uniform Motion (constant velocity, a = 0)
speed v = x/t
speed is the magnitude of the velocity vector
(a vector is a quantity w/ both magnitude and direction)
a parked car
skater coasting on a rink
a car moving up a hill at steady speed and
What happens when nothing is pushing her along?
Why does she continue to coast?
• Mass of skater (measured in kg) is
a measure of inertia (resistance to
a change in motion).
• Anything with mass will continue in
its current state of linear motion
Newton’s 1st Law of Motion:
An object that is not subject to
any outside forces moves at constant
velocity, covering equal distances in
equal times, along a straight-line path.
Coasting on Ice
Skates reduce friction, making
property of ‘inertia’ more pronounced
Everyday Demonstrations of Newton’s
First Law (Law of Inertia)
• Blood rushes from your head to your feet
while quickly stopping when riding on a
descending elevator.
• The head of a hammer can be tightened
onto the wooden handle by banging the
bottom of the handle against a hard
• To dislodge ketchup from the bottom of a
ketchup bottle, you shake it.
• Headrests are placed in cars to prevent
whiplash injuries during rear-end collisions.
2. Uniformly Accelerated Motion (a ≠ 0)
a = v/ t = (v-vo)/t …or v = vo + at
x = xo + vot + (1/2) a t2
These two equations help you predict motion of
uniformly accelerated objects.
But why will an object accelerate in the first place, and
change its velocity ?
Newton’s 2nd Law of Motion
A force F exerted on an object is equal to the product of
the object’s mass and its acceleration. The acceleration is in the
same direction as the applied force.
Examples of uniformly accelerated motion:
• your car when you drive it in a straight line
with the accelerator pressed constantly
• a ball falling under the influence of gravity
(freely-falling objects)
Question: A car moving to the right begins to apply its brakes
constantly. Where is the velocity pointing ?
Where is the acceleration pointing ?
Answer: v(t) pointing to the right, a(t) pointing to the left, opposite
to the motion, causing the car to decelerate.
Now: Let’s take a closer look at the case of a
ball dropped from rest.
How does a dropped ball ‘fall’? (Describe a(t), v(t), x(t))
We use v = vo + at and x = xo + vot + (1/2) a t2
where a = constant, = -g = - 9.8 m/s2 , downwards
We can simplify matters by choosing x0 (the initial position = 0);
v = vo – gt
and x = - (1/2)gt2
We can use these to simulate a ball dropping, finding its x and v
at different times:
0 secs
0 m/s
- 9.8 m/s2
1 secs
- 4.9 m
- 9.8 m/s
- 9.8 m/s2
2 secs
- 19.6 m
-19.6 m/s
- 9.8 m/s2
3 secs
- 44.1 m
- 29.4 m/s
- 9.8 m/s2
Free-fall or ‘falling’ - a special type of motion
in which the only force acting upon an object is
Weight (force of gravity)
Neglecting air friction, all objects will fall with the same
acceleration, regardless of their mass. Why ?
A Closer Look:
Big Rock vs Little Rock, which ‘falls faster’?
Assuming no air friction. both fall at same rate
Some Exercises
1. Bungee-Jumping Amusement
How long should the rope
be for a 5 – second free fall
experience ?
We know initial velocity vo = 0 and elapsed time t = 5 s
How to find final v ?
v = vo –g  t = 0 – (10 m/s2)(5 s) = - 49 m, or 49 meters down
2. You throw a ball up with initial speed of 10 m/s.
a. How high does it climb ? We know vo = 10m/s and v = 0
v = vo –g  t or 0 = 10 m/s – (10m/s2) t or t = 1 sec
But x = vot – (1/2)gt2 = x = (10m/s)(1s) – (1/2)(-10m/s2)(1s)2
or x = 5 meters
How fast does it come back at you? (discussed in class)